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Chapter 6
The Normal Distribution
Section 6-1
Normal Distributions
Learning Target
• IWBAT identify distributions as symmetric or
skewed.
• IWBAT identify the properties of a normal
distribution.
Symmetric or Skewed
• Symmetric – when the values are evenly
distributed about the mean (mean, median, and
mode are centrally located)
• Negatively (left-skewed) – majority of the data
values fall to the right of the mean (mean,
median, and mode are located in this order)
• Positively (right-skewed) – majority of the data
values fall to the left of the mean (mode,
median, and mean are located in this order)
Normal Distribution
A normal distribution is a
continuous, symmetric
bell-shaped distribution of
a variable.
Properties
1. Bell-shaped
2. Mean, median, and mode are
equal and located at the center
3. Unimodal
4. Symmetrical
5. Continuous
6. Curve never touches the x-axis
7. Total area under the curve is equal
to 1 or 100%
8. Area that lies within 1 st. dev. from
the mean is approx. .68 or 68%, 2
st. dev. is approx. .95 or 95%, and 3
st. dev. is approx. .997 or 99.7%
Standard Normal Distribution
• A normal distribution that has a mean of 0
and a standard deviation of 1.
• Uses z-scores to determine the number of
deviations a value is from the mean.
• Formula for z-score
» Z= (value – mean)/ standard deviation
• Use Table E (handout) – the chart ranges from
-3.49 to 3.49 and gives area to four decimal
places
How to find the area under the curve
Step 1: Draw the normal distribution curve and
shade the area.
Step 2: Find the appropriate figure in the
Procedure Table (next slide) and follow the
directions
Examples
• Find the area to the left of z = 1.99
• Find the area to the right of z = -1.16
• Find the area between z = 1.68 and z = -1.37
• Solutions
– 0.9767
– 1.0000 – 0.1230 = 0.8770
– 0.9535 – 0.0853 = 0.8682
Examples using Probability
• Find the probability of each
– P(0 < z < 2.32)
– P(z < 1.65)
– P(z > 1.91)
–
–
–
–
Solutions
0.9898 – 0.5000 = 0.4898 or 48.98%
0.9505 or 95.05%
1.0000 – 0.9719 = 0.0281 or 2.81%
Examples (finding z-score given area)
• Find the z value such that the area under the
standard normal distribution curve between 0
and the z value is 0.2123.
• Solution: add .5000 to 0.2123 because of the
area to the left of 0. Then find the value in the
chart and locate the z value that corresponds
with 0.7123, which is 0.56.
Exercises 6-1
1 – 47 odd
Section 6-2
Applications of the Normal
Distribution
Learning Target
• IWBAT find probabilities for a normally
distributed variable by transforming it into a
standard normal variable.
Steps to Application Problems
𝑣𝑎𝑙𝑢𝑒 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑋−𝜇
𝜎
1. Given the formula 𝑧 =
=
find the z-score that corresponds to the given
information in the problem.
2. Once you have found the z-score determine
whether you are finding the area to the left,
to the right, or in between.
3. Find the area.
Example 1
• A survey for the National Retail Federation found that
women spend on average $146.21 for the Christmas
holidays. Assume the standard deviation is $29.44. Find
the percentage of women who spend less than $160.00.
Assume the variable is normally distributed.
• Solution:
𝑋 − 𝜇 160 − 146.21
𝑧=
=
= 0.47
𝜎
29.44
We now find the area to the left because the problem states less than
$160.
Find the area to the left directly from the Table and you get 0.6808 or
68.08%.
Example 2
Each month a household generates of average of
about 28 pounds of newspaper for garbage or
recycling. Assume the standard deviation is 2
pounds. If a household is selected at random,
find the probability of it generating
a. Between 27 and 31 pounds per month
b. More than 30.2 pounds
Solution to Example 2
a. 𝑧 =
𝑋−𝜇
𝜎
=
27−28
2
= −0.5
𝑋 − 𝜇 31 − 28
𝑧=
=
= 1.5
𝜎
2
Find -0.5 < z < 1.5 Area = 0.9332 – 0.3085 = 0.6247
b. 𝑧 =
𝑋−𝜇
𝜎
Find z > 1.1
=
30.2−28
2
= 1.1
Area = 1 – 0.8643 = 0.1357
Example 3 (more than one is selected)
The American Automobile Association reports
that the average time it takes to respond to an
emergency call is 25 minutes. Assume the
variable is normally distributed and the standard
deviation is 4.5 minutes. If 80 calls are randomly
selected, approximately how many will be
responded to in less than 15 minutes?
Solution to Example 3
𝑋 − 𝜇 15 − 25
𝑧=
=
= −2.22
𝜎
4.5
Find z < -2.22
Area = 0.0132
To find how many calls will be made in less than
15 minutes, multiply the sample size (80) by
0.0132 to get 1.056. Hence, 1.056, or
approximately 1, call will be responded to in
under 15 minutes.
Find Data Values Given Specific
Probabilities
Steps
1. Given the probability, determine the z-score
you will need to use.
𝑋−𝜇
𝜎
2. Using the formula 𝑧 =
substitute in the
values for 𝑧, 𝜇, 𝜎.
3. Solve for X.
Note: You can also use the following formula
𝑋 =𝑧∙𝜎+𝜇
Example
To qualify for a police academy, candidates must
score in the top 10% on a general abilities test.
The test has a mean of 200 and a standard
deviation of 20. Find the lowest possible score
to qualify. Assume the test scores are normally
distributed.
Solution
- Find the z-score that corresponds with 10% or
0.10 to the right of the z-score.
- Substitute the 3 values into the formula. 𝑧 =
𝑋−𝜇
𝜎
→ 1.28 =
𝑋−200
20
→ 𝑥 = 225.60 𝑜𝑟 226
Exercises 6-2
Problems 1 – 12
Then
Problems 14 – 38 even